show that $|x|^{-(n-2)}$ is harmonic function

Question

let $f:\mathbb{R}^n \to \mathbb{R}$, and $f: \mathbf{x} \mapsto \Vert\mathbf{x}\Vert^{-(n-2)}$ show that $f$ is harmonic. I tried to take derivative by $x_1$ twice, and I got this result: $1/4 (-1 + 4 n^2) (x + x_{2...n})^{(-3/2 - n)}$ this of course doesn't sum up to zero.

where am I wrong? thanks

It seems that K.defaoite has vandalised the question: $\Vert\mathbf{x}\Vert^{-n}$ should be $\Vert\mathbf{x}\Vert^{-(n-2)}$. — TonyK, Jun 18 '20 at 13:04
I am not sure if I understand your expression for the derivative you obtain, but have you taken into account the chain rule? — Rik93, Jun 18 '20 at 13:18

score 3 · Answer 1 · answered Jun 18 '20 at 14:02

One trick here is to note that $f$ is invariant under rotations, so we might get some mileage out of using polar coordinates. In polar coordinates, $$f(r,\theta_1,\ldots,\theta_{n-1}) = r^{2-n}$$ and the Laplacian has the expression $$\Delta = \frac{1}{r^{n-1}}\frac{\partial}{\partial r}\bigg(r^{n-1}\frac{\partial}{\partial r}\bigg) + r^{-2}\bigg(\mbox{angular derivatives}\bigg)$$

As $f$ depends only on $r$, the angular derivatives vanish, so $$\begin{align*} \Delta f &= \frac{1}{r^{n-1}}\frac{\partial}{\partial r}\bigg( r^{n-1} \frac{\partial}{\partial r} r^{2-n}\bigg) \\ &= \frac{1}{r^{n-1}}\frac{\partial}{\partial r}\bigg(r^{n-1}(-n+2)r^{1-n}\bigg) \\ &= \frac{2-n}{r^{n-1}}\frac{\partial}{\partial r}(1)\\ &= 0 \end{align*}$$

score 2 · Accepted Answer · answered Jun 18 '20 at 14:03

Given $f(\mathbf{x})=\Vert\mathbf{x}\Vert^{-(n-2)}$ we need to show that $$\nabla^2 f =\sum ^{n}_{i=1}\frac{\partial^2 f}{\partial x_{i}^{2}} =0$$ Let's work out the first derivative: $$\frac{\partial }{\partial x_{i}} \| \mathbf{x} \| ^{-n+2} =( 2-n) \| \mathbf{x} \| ^{-n+1}\frac{\partial \| \mathbf{x} \| }{\partial x_{i}}$$ Recalling that $\| \mathbf{x} \| =\sqrt{\sum\nolimits ^{n}_{i=1} x_{i}^{2}}$, we can see that $$\frac{\partial }{\partial x_{i}} \| \mathbf{x} \| ^{-n+2}=( 2-n) \| \mathbf{x} \| ^{-n} x_{i}$$ Now, $$\frac{\partial^2 f}{\partial x_{i}^{2}}=\frac{\partial}{\partial x_i}\left( ( 2-n) \| \mathbf{x} \| ^{-n} x_{i}\right)$$ Using the product rule, $$\frac{\partial}{\partial x_i}\left( ( 2-n) \| \mathbf{x} \| ^{-n} x_{i}\right)=( 2-n)\left( -n\| \mathbf{x} \| ^{-n-2} x_{i}^{2} +\| \mathbf{x} \| ^{-n}\right)$$ $$=( 2-n) \| \mathbf{x} \| ^{-n-2}\left( -nx_{i}^{2} +\| \mathbf{x} \| ^{2}\right)$$ Thus $$\nabla^2 f = ( 2-n) \| \mathbf{x} \| ^{-n-2}\sum ^{n}_{i=1}\left( -nx_{i}^{2} +\| \mathbf{x} \| ^{2}\right)$$ $$=( 2-n) \| \mathbf{x} \| ^{-n-2}(-n\Vert\mathbf{x}\Vert^2+n\Vert\mathbf{x}\Vert^2)=0.$$

show that $|x|^{-(n-2)}$ is harmonic function

2 Answers2